/*Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
Given target = 3, return true.
*/
#if 0
class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();
        int row = m-1, col = 0;
        while(row>=0 && col<n) {
            if (matrix[row][col] == target) return true;
            else if (matrix[row][col] < target) {
                // move right
                col++;
            } else {
                // move up
                row--;
            }
        }
        return false;
    }
};
#endif

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        const size_t  m = matrix.size();
        const size_t n = matrix[0].size();
        int first = 0;
        int last = m * n - 1;
        while (first <= last) {
            int mid = first + (last - first) / 2;
            int value = matrix[mid / n][mid % n];
            if (value == target) {
                return true;
            } else if (value < target) {
                first = mid + 1;
            } else {
                last = mid - 1;
            }
        }
        return false;
    }
};
